hagman
Posts:
1,923
Registered:
1/29/05


Re: Proof 0.999... is not equal to one.
Posted:
May 31, 2007 3:32 PM


On 31 Mai, 14:04, chaja...@mail.com wrote: > This has been a very educational evening for me. I would still like to > point out that no one has refuted my proof itself or pointed to > specific error or logical overstepping it may contain. > > http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.doc > > http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.odt > > http://www17.brinkster.com/chajadan/Math/Proofs/Proof1.txt > > > Charles J. Daniels
OK, here are some notes:
Near the bottom of page 1 you write: Let S be the set of all real numbers in the interval [0,1] Let T be the set of all real numbers in the interval (1,0]
If you applied an operator '+' to these two sets that sums the elements of both sets, you would get: S + T = 1
You have never defined this operator '+' for intervals. The standard use (derived from the '+' for numbers) is to take the set of all possible sums x+y where x is in S and y is in T. However, this would result in S + T = (1,1].
Most of the text is so very handwavingly informal that it is not even wrong.
I found one portion, however, near the top of page 2, that sounds mathematically correct to me:
Let x = 0.999...
x = 0.999... {1} Multiply both sides by 10: 10x = 9.999... {2} Subtract line {1} from line {2}: 9x = 9 {3} Divide both sides by 9: x = 1
Since x = 0.999... and x = 1, it must be the case that 1 = 0.999...
Later, your arguments, especially on page 4 are totally circular.
Page 5 shows that you seem to grasp that there is a necessity to *define* the meaning of infinite sums since things can go wrong when rearranging series that are not absolutely converging.
To save my own sanity, I scrolled ahead until I found this example of downright nonsense:
However, defining it as a limit is all but directly saying that it may not be exactly equal to one.
Later on
A Nested Set theorem says that a sequence of nested closed sets that become arbitrarily small eventually contain only a single real number. Then they show something like this:
[0, 1], [0.9, 1], [0.99, 1], ..., [0.999..., 1]
But what about the closed set [1]? How could we have felt sure 0.999... would be in this set as well? The nested set theorem only guaranteed you'd get to a single value.
Indeed, you'd get only a single value AND you'd get 0.999... AND you'd get 1. This implies 0.999...=1.
Maybe, the problem is that you have never defined what "0.999..." means. In this thread you claimed that it should be the sum of all elements of the set S where S contains all numbers of the form 9/10^n for n>=0. So the problem is shifted to defining what the sum of all elements of an infinite set of positive rational numbers is. Two cases arise: a) For every integer M, we can find a finite subset of S such that the (clearly defined) sum of these finitely many elements is bigger than M. In this case, there is no hope of defining the sum of S or it is said to diverge. b) For some M, it is impossible to get beyond M with a finite subset. We might be able to define a sum here. In fact, *I* can make such a definition in a consistent manner.
One can show that your set S is of type b). Thus one can hope to define sum(S) and in fact whatever you claim 0.999... to be is what you define as sum in this case.
If r is a positive rational number, one can define the set T containing all elements of S multiplied by r, i.e. the numbers r*9/10^n.
It should definitely be true that, whenever either sum(S) or sum(T) is defined so is the other and the equality sum(T) = r * sum(S) holds, shouldn't it?
If s is a rational number not contained in S, we can define the set U = S u {s} containing all elements of S and also s. It should definitely be true that, whenever either sum(S) or sum(U) is defined so is the other and the equality sum(U) = s + sum(S) holds, shouldn't it?
With your set S defined as above, I choose r=10 to form T and I choose s=9 to form U. If you look at it closely, you will note that T=U holds. It should definitely be true that the equality sum(U) = sum(T) holds in this case, shouldn't it?
But then r * sum(S) = sum(T) = sum(U) = s + sum(S) i.e. 10 * sum(S) = 9 + sum(S).
In my math, this can be transformed to sum(S) = 1. Your mileage my vary, but then you have not given adequate definitions of '+', '*', '='.
hagman

